Proof Verification: If a vector space V is a union of k subspaces of its subspaces, prove that V is equal to one of its subspace.

Question

Suppose the subspaces are S1,S2,S3.....Sk

Take a from S1, b from S2....k from Sk

then a+b+....k belongs to V and union of all subspaces.

By the definition of union, a+b+....k must belong to atleast one of the subspace which will make it equal to V.

Is this proof correct? Or am I missing something?

Your phrasing is quite unclear, please try to be more precise. Also, the statement itself clearly fails for finite dimensional vector spaces over finite fields, which are indeed unions of some proper subspaces. — A. Pongrácz, Sep 01 '18 at 20:35
This is not clear. What does "its $k$ subspaces" mean? Vector spaces can have lots and lots of subspaces. — lulu, Sep 01 '18 at 20:36
Of its some k subspaces. So, we have to prove that V is equal to any one of the k subspaces we have selected. — Avinash Bhawnani, Sep 01 '18 at 20:37
@AvinashBhawnani That still isn't clear. Vector spaces over what? If the underlying field is finite the claim is clearly false, for example. — lulu, Sep 01 '18 at 20:40
@ArnaudMortier: It can be true over finite fields under cardinality conditions. — Bernard, Sep 01 '18 at 20:44
the field is not mentioned to me right now, but considering that we have not yet been taught vector spaces over finite fields, I assume we have to take infinite field only. Also, I have read fields over {0,1}, would also appreciate a counter example in a finite field. — Avinash Bhawnani, Sep 01 '18 at 20:44
Okay, I got the example for finite field from another question. Is this proof valid if the field is infinite? — Avinash Bhawnani, Sep 01 '18 at 20:49

score 2 · Accepted Answer · answered Sep 01 '18 at 20:54

2

Actually, if a vector space over a field $F$ is the union of $k$ proper subspaces, then $|F|<k$. So if the field is infinite, one of the subspaces is equal to $V$. This result is known as the avoidance lemma for subspaces.

For a proof you can take a look at my answer to this closely related question

answered Sep 01 '18 at 20:54

Bernard

175,478

I didn't understand why at most 1 element? – Avinash Bhawnani Sep 01 '18 at 21:05
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Oh! In the proof? It is because if $v+ku$ and $v+k'u\in W_i$ for some $k\ne k'$; as W_i is a subspace, we deduce that $(v+ku)-(v+k'u)=(k-k')u \in W_i$, and as $k-k'\ne 0$, this implies $u\in W_i$, contrary the the hypothesis. – Bernard Sep 01 '18 at 21:12
Thank you! For infinite elements, is my proof correct? (Given that written in Mathematical notations, don't know how to write here) – Avinash Bhawnani Sep 01 '18 at 21:18
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No it is not correct, because you've forgotten that when you change the value of some of $a,b, \dots,k$, the sum $a+b+\dots+k$ lies in one of the subspaces, but not necessarily in the same subspace as for the previous values. For the formatting with Mathjax, it is very close to the formatting of maths formulæ with LaTeX, if you know it. – Bernard Sep 01 '18 at 21:28
I actually just tried to extend the proof of "if union of two subspaces is a subspace then one must contain the other"
If a+b+c....+k belongs in V, then it must contain in one of its subspaces. Suppose S3.

So, now a+b+....k belongs in S3. But we only took c from S3, so from the subspace property, a,b,d...,k must belong in S3. Which will make S3 =V?
– Avinash Bhawnani Sep 01 '18 at 21:33
I can't follow you. You should edit you post and give details, so we can try to see what's correct and what's wrong. – Bernard Sep 01 '18 at 21:43
Oh! I got my mistake! I was writing the proof of mine to edit the post, and I understood your comment while writing. Thank you again! :) – Avinash Bhawnani Sep 01 '18 at 21:50

Proof Verification: If a vector space V is a union of k subspaces of its subspaces, prove that V is equal to one of its subspace.

1 Answers1